ENC3250 Solutions pg 37

ENC3250 Solutions pg 37 - 8.7 a. The multiplication of M(X)...

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Unformatted text preview: 8.7 a. The multiplication of M(X) by X16 corresponds to shifting M(X) 16 places and thus providing the space for a 16-bit FCS. The addition of XkL(X) to X16M(X) inverts the first 16 bits of G(X) (one's complements). The addition of L(X) to R(X) inverts all of the bits of R(X). b. The HDLC standard provides the following explanation. The addition of XK L(X) corresponds to a value of all ones. This addition protects against the obliteration of leading flags, which may be non-detectable if the initial remainder is zero. The addition of L(X) to R(X) ensures that the received, errorfree message will result in a unique, non-zero remainder at the receiver. The non-zero remainder protects against the potential non-detectability of the obliteration of trailing flags. c. The implementation is the same as that shown in Solution 3b, with the following strategy. At both transmitter and receiver, the initial content of the register is preset to all ones. The final remainder, if there are no errors, will be 0001 1101 0000 1111. 8.8 a. For simplicity, we do not show the switches. b. C4 C3 C2 C1 C0 C 4 ⊕ C3 C 4 ⊕ C1 C 4 ⊕ I Input 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 0 0 0 0 0 1 0 1 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 1 1 0 1 0 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 0 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 – 0 1 0 1 1 1 1 0 – 1 0 1 1 0 1 0 1 – 1 1 0 0 1 0 1 1 – 0 0 1 1 1 0 1 0 – 0 1 1 1 0 c. The partial results from the long division show up in the shift register, as indicated by the shaded portions of the preceding table. Compare to long division example in Section 8.1. d. Five additional steps are required to produce the result. -37- ...
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